Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

I need to find the area of the yellow part of the arc given the a, b , start and end angle of the sector points

Also the ellipse is centered at the origin

How to find the area of the yellow part?

share|cite|improve this question
If you stretch the figure vertically by a factor of $a/b$ (keeping the $x$-axis fixed), you'll get a circle of radius $a$. The stretched yellow area will be $a/b$-times the original yellow area. The (slightly-)tricky part is determining the starting and ending angles in the stretched figure, but once you have them, the area of a circular segment is straightforward to compute. – Blue Aug 3 '13 at 4:37
I got this link: but don't know hat is the r(delta) in the formula – andikat dennis Aug 3 '13 at 5:11

If we are allowed to use calculus, the eqaution of the ellipse, $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and $\alpha,\beta$ be the known angles

From the defintion of parametric angle,

We get, $\tan\alpha=\frac{b\sin\phi}{a\cos\phi}$ and $\tan\beta=\frac{b\sin\delta}{a\cos\delta}$ where $\phi,\delta$ are the parametric angles

Consequently,the area will be $$\left|\int_{a\cos\phi}^{a\cos\delta}ydx\right|=\left|\int_{a\cos\phi}^{a\cos\delta}b\left(\sqrt{1-\frac{x^2}{a^2}}\right)dx\right|=\frac ba\left|\int_{a\cos\phi}^{a\cos\delta}\sqrt{a^2-x^2}dx\right|$$

share|cite|improve this answer
@AndréNicolas, not sure if you have seen the edited version? We need $\int_{x_1}^{x_2} ydx$ where $x_1=a\cos\phi$ where $\tan\phi=\frac ab\tan\alpha,$ where $\alpha$ is known – lab bhattacharjee Aug 3 '13 at 4:42

A calculus-free derivation:

Consider the analagous figure drawn for a unit circle. We find the area (assuming an angle is given as $\theta$) as $$ A = \frac12(\theta-\sin\theta) $$

Stretch the graph left-right by a factor of $a$, and stretch it up-down by a factor of $b$. Having stretched the region with the rest of the picture, we can deduce that the new area will be $$ A = \frac{ab}{2}(\theta-\sin\theta) $$ Where $\theta$ is still the angle of our squished ellipse. To make this a complete formula, we must find an expression for $\theta$ given an elliptical angle.

In fact, if we are given an elliptical angle $\phi$ from the x-axis, we have $$ \theta = \arctan\left[\frac{a}b \tan\phi\right] = $$ Which gives us $$ A = \frac{ab}{2}\left(\arctan\left[\frac{a}b \tan\phi\right]-\sin\left(\arctan\left[\frac{a}b \tan\phi\right]\right)\right) $$ In the case that $\phi$ is not given as an angle from the x-axis, we can break $\phi$ into $\phi = \phi_1+\phi_2$, where $\phi_1$ is the part of the angle going clockwise from the x-axis, and $\phi_2$ is the counterclockwise part from the x-axis. We then have

$$ \theta = \arctan\left[\frac{a}b\phi_1\right] + \arctan\left[\frac{a}b\phi_2\right]\\ %\tan\theta= ab\frac{\tan\phi_1+\tan\phi_2}{1-a^2\tan\phi_1\tan\phi_2} $$ Which can be substituted as before.

This was more complicated than I expected it to be.

Please comment, edit, or let me know if there is anything I have left out that makes this answer less understandable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.